Question

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Answer 1

Ответ:

$A)\ x_1 = -2,\ x_2 =\dfrac{1}{5};\\B)\ x_{1/2} = б2;\\C)\ x = 3.$

Объяснение:

$A)\ -5x^2-9x+2 = 0\ (*(-1))\\5x^2+9x-2 = 0\\\boxed{D = b^2 -4ac}\\D = 9^2 - 4 * 5 *(-2) = 81 + 40 = 121\\\\\boxed{x_{1/2} = \dfrac{-bб\sqrt{D} }{2a}}\\\\x_{1/2} = \dfrac{-9б11}{10}\\\\x_1 = \dfrac{-9-11}{10} = -\dfrac{20}{10} = -2;\\\\x_2 = \dfrac{-9+11}{10} = \dfrac{2}{10} = \dfrac{1}{5}.$

$B)\ x^4+5x^2-36=0\\(x^2)^2 + 5x^2-36 = 0\\\boxed{x^2 = y}\\y^2 + 5y - 36 = 0\\\boxed{D = b^2-4ac}\\D = 5^2 - 4*1*(-36) = 25 + 144 = 169\\\\\boxed{y_{1/2} = \dfrac{-bб\sqrt{D} }{2a}}\\\\y_{1/2} = \dfrac{-5б13}{2}\\\\y_1 = \dfrac{-5-13}{2} = -\dfrac{18}{2} = -9;\\\\y_2 = \dfrac{-5+13}{2} = \dfrac{8}{2} = 4.\\Zamena:\\x^2 = -9\\x \in \varnothing\\\\x^2 = 4\\x = б\sqrt{4}\\x_{1/2} = б2$

$C)\ \dfrac{2}{x-4} + \dfrac{x}{x+4} = \dfrac{20-3x}{x^2-16}\\\\\dfrac{2}{x-4} + \dfrac{x}{x+4} - \dfrac{20-3x}{x^2-16} = 0\\\\\dfrac{2^{/x+4}}{x-4} + \dfrac{x^{/x-4}}{x+4} - \dfrac{20-3x}{(x-4)(x+4)} = 0\\\\\dfrac{2(x+4)+x(x-4)-(20-3x)}{(x-4)(x+4)} = 0\\ \\\dfrac{2x+8+x^2-4x-20+3x}{(x-4)(x+4)} = 0\\\\\dfrac{x^2+x-12}{(x-4)(x+4)} = 0\\-------\\ODZ:\\(x-4)\neq 0\\x\neq 4\\(x+4)\neq 0\\x\neq -4\\-------\\x^2+x-12 = 0\\\boxed{D = b^2-4ac}\\D = 1^2 - 4 *1*(-12) = 1 + 48 = 49$

$\boxed{x_{1/2} = \dfrac{-bб\sqrt{D} }{2a}}\\\\x_{1/2} = \dfrac{-1б7}{2}\\\\x_1 = \dfrac{-1-7}{2} = -\dfrac{8}{2} = -4 - Ne\ v\ ODZ;\\\\x_2 = \dfrac{-1+7}{2} = \dfrac{6}{2} = 3.$